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\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)

\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)

\(\Leftrightarrow-28x+37\ge12\)

\(\Leftrightarrow-28x\ge12-37\)

\(\Leftrightarrow-28x\ge-25\)

\(\Leftrightarrow x\le\dfrac{25}{28}\)

Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)

b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)

\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)

\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)

\(\Leftrightarrow-6x\ge30\)

\(\Leftrightarrow x\le-5\)

Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)

\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)

\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)

\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)

\(\Leftrightarrow-11x+37< 0\)

\(\Leftrightarrow-11x< -37\)

\(\Leftrightarrow x>\dfrac{37}{11}\)

vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)

a) ta có : \(3x-5>2\left(x-1\right)+x\Leftrightarrow3x-5>2x-2+x\)

\(\Leftrightarrow-5>-2\left(vôlí\right)\) \(\Rightarrow x\in\varnothing\)

b) ta có : \(\left(x+2\right)^2-\left(x-2\right)^2>8x-2\Leftrightarrow8x>8x-2\)

\(\Leftrightarrow0>-2\left(đúng\forall x\right)\) \(\Rightarrow x\in R\)

c) ta có : \(3\left(4x+1\right)-2\left(5x+2\right)\ge8x-2\)

\(\Leftrightarrow12x+3-10x-4\ge8x-2\Leftrightarrow-6x\ge-1\Leftrightarrow x\le\dfrac{1}{6}\)

d) ta có : \(2x^2+2x+1-\dfrac{15\left(x+1\right)}{2}\le2x\left(x+1\right)\)

\(\Leftrightarrow2x^2+2x+\dfrac{2-15x-15}{2}\le2x^2+2x\)

\(\Rightarrow\dfrac{-15x-13}{2}\le0\Leftrightarrow-15x-13\le0\Leftrightarrow x\ge\dfrac{-13}{15}\)

a) \({2^x} > 16 \Leftrightarrow {2^x} > {2^4} \Leftrightarrow x > 4\) (do \(2 > 1\)) .

b) \(0,{1^x} \le 0,001 \Leftrightarrow 0,{1^x} \le 0,{1^3} \Leftrightarrow x \ge 3\) (do \(0 < 0,1 < 1\)).

c) \({\left( {\frac{1}{5}} \right)^{x - 2}} \ge {\left( {\frac{1}{{25}}} \right)^x} \Leftrightarrow {\left( {\frac{1}{5}} \right)^{x - 2}} \ge {\left( {{{\left( {\frac{1}{5}} \right)}^2}} \right)^x} \Leftrightarrow {\left( {\frac{1}{5}} \right)^{x - 2}} \ge {\left( {\frac{1}{5}} \right)^{2x}} \Leftrightarrow x - 2 \le 2{\rm{x}}\) (do \(0 < \frac{1}{5} < 1\))

\( \Leftrightarrow x \ge  - 2\).

\(\left(x-4\right).\left(x+4\right)\ge\left(x+3\right)^2+5\)

\(\Rightarrow x^2-16\ge x^2+6x+9+5\)

\(\Rightarrow x^2-16\ge x^2+6x+14\)

\(\Rightarrow-30\ge6x\Rightarrow-5\ge x\)

Vậy...

a, ĐK: \(x-2>0\Rightarrow x>2\)

\(log_2\left(x-2\right)< 2\\ \Leftrightarrow x-2< 4\\ \Leftrightarrow x< 6\)

Kết hợp với ĐKXĐ, ta được: \(2< x< 6\)

b, ĐK: \(2x-1>0\Leftrightarrow x>\dfrac{1}{2}\)

\(log\left(x+1\right)\ge log\left(2x-1\right)\\ \Leftrightarrow x+1\ge2x-1\\ \Leftrightarrow x\le2\)

Kết hợp với ĐKXĐ, ta được: \(\dfrac{1}{2}< x\le2\)

ĐKXĐ:\(\left\{{}\begin{matrix}x\ne1\\x\ne2\\x\ne7\end{matrix}\right.\)

\(\dfrac{2\left(x-4\right)}{\left(x-1\right)\left(x-7\right)}\ge\dfrac{1}{x-2}\\ \Leftrightarrow\dfrac{2x-8}{x^2-8x+7}\ge\dfrac{1}{x-2}\\ \Leftrightarrow\left(2x-8\right)\left(x-2\right)\ge x^2-8x+7\)

\(\Leftrightarrow2x^2-12x+16\ge x^2-8x+7\\ \Leftrightarrow x^2-4x+9\ge0\left(luôn.đúng\right)\)

Nhận xét rằng \(\sqrt{5}-2=\left(\sqrt{5}-2\right)^{-1}\)

Do đó bất phương trình có thể viết thành :

\(\left(\sqrt{5}-2\right)^{x+1}\ge\left[\left(\left(\sqrt{5}-2\right)^{-1}\right)\right]^{x-3}=\left(\left(\sqrt{5}-2\right)^{3-x}\right)\)

\(\Leftrightarrow x+1\ge3-x\)

\(\Leftrightarrow x\ge1\)

Vậy tập nghiệm của phương trình là :

\(D\left(1;+\infty\right)\)